JEE Main 2020 · 2 September · evening · Single correct · +4 / −1

Let E C denote the complement of an event E. Let E₁ , E₂ and E₃ be any pairwise independent events with P(E₁) > 0
and P(E₁ ∩ E₂ ∩ E₃) = 0.
Then P(E₂^C ∩ E₃^C/E₁) is equal to :

  1. (a) P( E₃^C ) - P(E₂)
  2. (b) P( E₂^C ) + P(E₃)
  3. (c) P( E₃^C ) - P( E₂^C )
  4. (d) P(E₃) - P( E₂^C )
Answer and solution

Answer (a) P( E₃^C ) - P(E₂)

Solution

Maths sir · 4 steps

Turn the two complements into one complement of a union, then subtract the overlaps inside E₁.

  1. De Morgan first. E₂^C ∩ E₃^C = (E₂ ∪ E₃)^C. So, inside E₁, we want the part of E₁ that lies in neither E₂ nor E₃.

    P(E₁ ∩ E₂^C ∩ E₃^C) = P(E₁) - P(E₁ ∩ E₂) - P(E₁ ∩ E₃) + P(E₁ ∩ E₂ ∩ E₃)

  2. Use the given facts. Pairwise independence gives P(E₁ ∩ E₂) = P(E₁)P(E₂) and P(E₁ ∩ E₃) = P(E₁)P(E₃). The triple intersection is given as 0.
  3. Divide by P(E1). Since P(E₁) > 0, the conditional probability is P(E₁)[1 - P(E₂) - P(E₃)]/P(E₁) = 1 - P(E₃) - P(E₂).
  4. So. 1 - P(E₃) = P(E₃^C), so the answer is P(E₃^C) - P(E₂), which is option (a).

The idea to keep. For "neither A nor B", write it as the complement of A ∪ B and use inclusion-exclusion. Pairwise independence turns every two-event overlap into a product.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Probability — Advanced

Asked about 100 times in JEE Main since 2020. This question is from 2020.